If `1 1/2` moles of oxygen combine with Al to form `Al_(2)O_(3)` the weight of Al used in the reaction is (Al=27)

Options

A. `27 g`
B. `54 g`
C. `40.5 g`
D. `81 g`

Explanation

To solve the problem, we need to determine the weight of aluminum (Al) used when 1.5 moles of oxygen (O₂) combine to form aluminum oxide (Al₂O₃). Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Reaction
The balanced chemical equation for the formation of aluminum oxide is:
\[ 4Al + 3O_2 \rightarrow 2Al_2O_3 \]
From this equation, we can see the molar ratio of aluminum to oxygen.

### Step 2: Determine the Molar Ratio
From the balanced equation:
- 3 moles of O₂ react with 4 moles of Al.

### Step 3: Calculate Moles of Aluminum
Given that we have 1.5 moles of O₂, we can use the molar ratio to find the moles of aluminum required:
\[ \text{Moles of Al} = \left(\frac{4 \text{ moles Al}}{3 \text{ moles O}_2}\right) \times \text{Moles of O}_2 \]
Substituting the value:
\[ \text{Moles of Al} = \left(\frac{4}{3}\right) \times 1.5 \]
Calculating this gives:
\[ \text{Moles of Al} = \frac{4 \times 1.5}{3} = 2 \text{ moles of Al} \]

### Step 4: Calculate the Mass of Aluminum
Now that we know we need 2 moles of aluminum, we can calculate the mass using the molar mass of aluminum (which is 27 g/mol):
\[ \text{Mass of Al} = \text{Moles of Al} \times \text{Molar Mass of Al} \]
Substituting the values:
\[ \text{Mass of Al} = 2 \text{ moles} \times 27 \text{ g/mol} = 54 \text{ grams} \]

### Final Answer
The weight of aluminum used in the reaction is **54 grams**.

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